44.

45.

+

+

(x + 2)2 - y2 ' (x +{y)2 - 22 +;

xy- y2

22 - (x − y)3 ̧
(y+2)2-22

x2 - (y-2x)2, y2 - (2 z − x)2, 4 z2 — (x − y)2.

(2 z + x)2 - y2 + (x + y)2 - 422 + (y+22)2-x2

(xy) (y-2)+(y − z) (≈ − x) + (≈ − x) (x − y) ̧
x(≈ − x) + y(x − y) + z(y − z)

c-a-b
(c− a)(c—b)

x2-(2y-3 z)2, 4y2 - (3z-x)2, 9z2-(x-2y)2

b + c

+

51.

52.

53.

54.

(2x+3y)2-16 22+

1

a

x2 + a2

+

a + x 1

16 z2-(2x-3y)2 4x2-(3y-4 z)2

(3y+4z)2 - 4x2 + (4x+2x)2-9 y2

a2 + x2 α a2 + x2

(x+a)(x+b)−(y+a)(y+b) _ (x−a) (y—b)−(x—b) (y—a)

CHAPTER XV.

FRACTIONAL AND LITERAL EQUATIONS.

163. In this chapter we propose to give a miscellaneous collection of equations. Some of these will serve as a useful exercise for revision of the methods already explained in previous chapters; but we also add others presenting more difficulty, the solution of which will often be facilitated by some special artifice.

The following examples worked in full will sufficiently illustrate the most useful methods.

11x+994 x - 44;

. 11 x 4x=- 4435299;

collecting terms and changing signs, 15 x = 495;

the minus sign before it. In fact it is equivalent to -(x-9), the line between the numerator and denominator having the same effect as a bracket.

This equation might be solved by clearing of fractions, but the work would be very laborious. The solution will be much simpli

(x −8) (x − 7) — (x − 5) (x − 10) – (x − 7) (x − 6) – (x − 4) (x − 9).

(x-10)(x-7)

=

(x-9)(x-6)

x2 — 15 x+56 — (x2 — 15 x+50) – x2−13x+42 − (x2−13x+36);

(x-10)(x-7)

6

=

(x − 10) (x − 7) ̄ ̄ ̄ (x − 9) (x − 6)

Hence, since the numerators are equal, the denominators must be equal; that is,

x-
1

- 13

х 6

4 x 55 x- - 6

We have 5+ 13-(2+) = 4+1 (1 + 2);

164. To solve equations whose coefficients are decimals we may express the decimals as common fractions, and proceed as before; but it is often found more simple to work entirely in decimals.

clearing of fractions, 24x+9-4x=68 - 27 x

transposing,

-

12;

29. .083(x-.625) = .09(x-.59375).

+

2 x 7 X

10x -8

X

6

x 4

.0625

30. (2x+1.5) (8x-2.25) = (2x-1.125) (8x+1.25).

1-1.4.7(x-1)
.2 + x .1–6

(.3 x − 2) (.3 x − 1) — 1 (.3 x − 2) = .4 x − 2.

.2x-1

LITERAL EQUATIONS.

165. In the equations we have discussed hitherto the coefficients have been numerical quantities, but equations often involve literal coefficients. [Art. 6.] These are sup posed to be known, and will appear in the solution.

Ex. 1. Solve (x + α) (x + b) − c (a + c) = (x − c)(x + c) + ab. Multiplying out, we have

whence

x2 + ax + bx+ ab — ac — c2 = x2 — c2 + ab;